RLC dynamics and electrical resonance
An ideal series RLC circuit. Connect electrical damping and resonance to the mechanical system you already know.
First, picture it
Does adding resistance change the final capacitor voltage or the way it gets there?
What the model says
A capacitor's voltage and an inductor's current cannot jump under ordinary finite inputs in the ideal model. They therefore make useful states. The capacitor relation is i = C vC′, while Kirchhoff's voltage law gives L i′ = vin−Ri−vC. Eliminating current yields LC vC″ + RC vC′ + vC = vin. The zero-initial-state capacitor-voltage transfer function is 1/(LCs²+RCs+1). Initial charge is a separate free-response contribution.
C vC′ = i; L i′ = vin − Ri − vC; E = ½C vC² + ½L i²Ideal voltage source, linear resistor, inductor and capacitor. There is no feedback controller, source current limit, parasitic resistance or breakdown model. RK4 at 1 ms integrates both states, source work and resistor loss; samples every 20 ms. E(t) − E(0) = source work − resistor loss. Large resonance voltages are mathematical predictions within these assumptions, not equipment ratings. A finite trace may still include startup transients.
Choose states that describe stored energy
A capacitor's voltage and an inductor's current cannot jump under ordinary finite inputs in the ideal model. They therefore make useful states. The capacitor relation is i = C vC′, while Kirchhoff's voltage law gives L i′ = vin−Ri−vC. Eliminating current yields LC vC″ + RC vC′ + vC = vin. The zero-initial-state capacitor-voltage transfer function is 1/(LCs²+RCs+1). Initial charge is a separate free-response contribution.
Divide the denominator by LC to identify ωn = 1/√(LC) and ζ = R√(C/L)/2. Critical damping occurs at R = 2√(L/C). For L=1 H and C=0.0625 F, ωn=4 rad/s and critical R=8 Ω. These intentionally slow, ideal values make the motion easy to inspect; they are not a recommended component selection for hardware.
Use charge for the mechanical analogy
Let q be capacitor charge, so q′=i and vC=q/C. The equation becomes L q″ + R q′ + q/C = vin. Compare m x″ + c x′ + kx = F: q↔x, i↔x′, L↔m, R↔c and 1/C↔k. This is a force–voltage analogy. Capacitor voltage is q/C, not charge itself, so a voltage-to-voltage plot has a different numerator from a force-to-displacement plot.
Stored energy is E = ½C vC² + ½L i². Differentiate and substitute the state equations: E′ = vin i−R i². The source can supply or absorb power depending on the signs of voltage and current; source work need not be positive at every instant. Resistance always dissipates nonnegative power. This balance is a useful independent check of a numerical simulation.
Resonance depends on the output you ask about
For positive R, series-current magnitude reaches its maximum at 1/√(LC). Capacitor-voltage magnitude instead peaks at ωn√(1−2ζ²) only when 0<ζ<1/√2. With ζ=0.125, the voltage peak is around 3.94 rad/s rather than 4. A measured output includes its own scaling and frequency dependence; do not label every maximum simply as the natural frequency.
A short sine experiment contains the forced response and the initial transient. With zero resistance, the transient does not decay; at exact natural frequency the nonzero sinusoidal input produces growing oscillations and no bounded steady-state amplitude. Compare the end of matched-duration recordings and explain the assumptions before reading a peak as a frequency-response gain.
Make it concrete
Pin R = 1 Ω for a 5 V step. Change only R to 8 Ω (critical), then 10 Ω. Compare capacitor voltage and current; keep L, C and initial conditions fixed.
Open the connected experimentMake a prediction before moving a control.
- PREDICTDoes adding resistance change the final capacitor voltage or the way it gets there?
- CHANGE ONE THINGPin R = 1 Ω for a 5 V step. Change only R to 8 Ω (critical), then 10 Ω. Compare capacitor voltage and current; keep L, C and initial conditions fixed.
- OBSERVEPin the baseline. Compare the same time and inspect the physical input as well as the output.
- EXPLAINFor this ideal series circuit, LC vC″ + RC vC′ + vC = vin. Positive resistance dissipates energy and the capacitor approaches the constant source voltage. R changes damping, not that DC gain. At R = 0, a step can ring forever; the existence of a DC equilibrium does not imply convergence.
For this ideal series circuit, LC vC″ + RC vC′ + vC = vin. Positive resistance dissipates energy and the capacitor approaches the constant source voltage. R changes damping, not that DC gain. At R = 0, a step can ring forever; the existence of a DC equilibrium does not imply convergence.
Check your understanding+
Does adding resistance change the final capacitor voltage or the way it gets there?
For this ideal series circuit, LC vC″ + RC vC′ + vC = vin. Positive resistance dissipates energy and the capacitor approaches the constant source voltage. R changes damping, not that DC gain. At R = 0, a step can ring forever; the existence of a DC equilibrium does not imply convergence.